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Question: An electric iron draws a current of \(15\,A\) from a \(220\,V\) supply. What is the cost of using ir...

An electric iron draws a current of 15A15\,A from a 220V220\,V supply. What is the cost of using iron for 30min30\,\min everyday for 15days15\,days if the cost of 1unit1\,unit is Rs.2Rs.\,2?
(A) Rs.49.5Rs.\,49.5
(B) Rs.60Rs.\,60
(C) Rs.40Rs.\,40
(D) Rs.10Rs.\,10

Explanation

Solution

The current which are used for households is expressed in the unit of kilowatt-hour. So, first determine the amount of power which the electric iron draws, then the power is multiplied with the hours of iron using the current, then we get the amount of current used by iron. After multiplying with the amount. Then the required cost can be determined.

Useful formula:
Electric power is equal to the product of the voltage and the current,
P=V×IP = V \times I
Where, PP is the power, VV is the voltage and II is the current,

Complete step by step solution:
Given that,
The current draws by iron, I=15AI = 15\,A
The voltage of the supply, V=220VV = 220\,V
Hours of using iron per day 30min\Rightarrow 30\,\min
Number of days using 15days \Rightarrow 15\,days
The amount of power used by iron,
P=V×I...............(1)P = V \times I\,...............\left( 1 \right)
Substituting the voltage and current value in the equation (1), then the equation (1) is written as,
P=220×15P = 220 \times 15
On multiplying the above equation, then the above equation is written as,
P=3300WP = 3300\,W
In other terms, the above equation is written as,
P=3.3kWP = 3.3\,kW
Time of iron used in one day,
30min\Rightarrow 30\,\min
0.5hr\Rightarrow 0.5\,hr
Number of days the iron used, 15days \Rightarrow 15\,days
The total time the iron used for 15days15\,days is 15×0.5hr \Rightarrow 15 \times 0.5\,hr
The total time iron used, t=7.5hrt = 7.5\,hr
So, the total unit of current used for 15days15\,days is,
P×t\Rightarrow P \times t
On substituting the power and time value in the above equation, then
3.3×7.5\Rightarrow 3.3 \times 7.5
On multiplying the above equation, then
24.75kWhr1\Rightarrow 24.75\,kWh{r^{ - 1}}
In other words, 24.75units24.75\,units of current is used by the iron for 15days15\,days.
The cost of 1unit1\,unit is Rs.2Rs.\,2, then the cost for 24.75units24.75\,units is Rs.2×24.75Rs.\,2 \times 24.75.
Then, the cost of using iron for 30min30\,\min everyday for 15days15\,days is Rs.49.5Rs.\,49.5.

Hence, the option (A) is correct.

Note: The current which are used for household is commonly expressed as unit, and this unit is expressed as kWhkWh. So, the time is converted from minutes to hours and then the total amount of hours is calculated by multiplying with the number of days. Then, the unit is determined. After multiplying the units with the amount, the total amount is determined.